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Ajax function is not returning any output

I am using this ajax function to get options for two dropdown via ajax but it is not returning any output. Hi this is my ajax function :

 <script>
$('#department').on('change',function(){
    var department = $(this).val();
    var course = $('#course').val();
    if(department){
        $.ajax({
            type:'POST',
            url:'ajaxData.php',
            dataType: 'json',
            cache: false,
            data:{department: department, course: course },
            success: function(data){
                $('#head_name').html(data.head_name);
                $('#email').html(data.email);
            }
        }); 
    }else{
        $('#head_name').html('<option value="">Select Department first</option>'); 
        $('#email').html('<option value="">Select Department first</option>'); 
    }
});
</script>

And this is my query code:

if(isset($_POST["department"]) && isset($_POST["course"])){
    //Get all courses data

    $query = $db->query("SELECT head_name, email FROM head WHERE course = '".$_POST['course']."' AND department = '".$_POST['department']."' ");

    //Count total number of rows
    $rowCount = $query->num_rows;

    //Display result list
    if($rowCount > 0){
        while($row = $query->fetch_assoc()){

            $temp = array('head_name' => '<option value="'.$row['head_name'].'">'.$row['head_name'].'</option>', 'email' => '<option value="'.$row['email'].'">'.$row['email'].'</option>' );
            echo json_encode($temp);
        }
    }else{
        $temp = array('head_name' => '<option value="">Not avaialble </option>', 'email' => '<option value="">Not avaialble </option>' );
        echo json_encode($temp);
    }
}

But I'm not getting output result from it, what I'm doing wrong ?

about 4 years ago · Santiago Trujillo
3 answers
Answer question

0

Your Ajax call expects JSON (dataType: 'json') but you do not return JSON via your PHP output because the default content type is text/html and not application/json. Add the appropriate Content-Type header before echo'ing your JSON output.

header('Content-Type: application/json');
echo json_encode(array('something' => 'else'));
about 4 years ago · Santiago Trujillo Report

0

you echo an array so you should access to data element as an array by index ..

eg for first element

       success: function(data){
            $('#head_name').html(data[0].head_name);
            $('#email').html(data[0].email);
        }

eventually try jsonParse data

       success: function(data){
             jsonData = JSON.parse(data);
            $('#head_name').jsonData(data[0].head_name);
            $('#email').jsonData(data[0].email);
        }

in your while you should use an array for store the values

        while($row = $query->fetch_assoc()){

        $temp[] = array('head_name' => '<option value="'.$row['head_name'].'">'.$row['head_name'].'</option>', 'email' => '<option value="'.$row['email'].'">'.$row['email'].'</option>' );
        echo json_encode($temp);
    }
about 4 years ago · Santiago Trujillo Report

0

I have just change the variable name of course and my function starts working , don't know why like this:

data:{ department: department, courseid: course},
about 4 years ago · Santiago Trujillo Report
Answer question
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